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  • 135. Candy

    There are N children standing in a line. Each child is assigned a rating value.

    You are giving candies to these children subjected to the following requirements:

    • Each child must have at least one candy.
    • Children with a higher rating get more candies than their neighbors.

    What is the minimum candies you must give?

    Example 1:

    Input: [1,0,2]
    Output: 5
    Explanation: You can allocate to the first, second and third child with 2, 1, 2 candies respectively.
    

    Example 2:

    Input: [1,2,2]
    Output: 4
    Explanation: You can allocate to the first, second and third child with 1, 2, 1 candies respectively.
                 The third child gets 1 candy because it satisfies the above two conditions.

    Approach #1: C++. [Using two array]

    class Solution {
    public:
        int candy(vector<int>& ratings) {
            int size = ratings.size();
            int ans = 0;
            vector<int> ltor(size, 1);
            vector<int> rtol(size, 1);
            
            for (int i = 1; i < size; ++i) {
                int j = size - i - 1;
                if (ratings[i] > ratings[i-1]) ltor[i] = ltor[i-1] + 1;
                if (ratings[j] > ratings[j+1]) rtol[j] = rtol[j+1] + 1;
            }
            
            for (int i = 0; i < size; ++i) 
                ans += max(ltor[i], rtol[i]);
            
            return ans;
        }
    };
    

      

    Approach #2: Java. [Using one array]

    public class Solution {
        public int candy(int[] ratings) {
            int[] candies = new int[ratings.length];
            Arrays.fill(candies, 1);
            for (int i = 1; i < ratings.length; i++) {
                if (ratings[i] > ratings[i - 1]) {
                    candies[i] = candies[i - 1] + 1;
                }
            }
            int sum = candies[ratings.length - 1];
            for (int i = ratings.length - 2; i >= 0; i--) {
                if (ratings[i] > ratings[i + 1]) {
                    candies[i] = Math.max(candies[i], candies[i + 1] + 1);
                }
                sum += candies[i];
            }
            return sum;
        }
    }
    

      

    Approach #3: Java. [Single Pass Approach with Constant Space]

    class Solution {
        public int count(int n) {
            return (n * (n + 1)) / 2;
        }
        
        public int candy(int[] ratings) {
            if (ratings.length <= 1) return ratings.length;
            int candies = 0;
            int up = 0;
            int down = 0;
            int old_slope = 0;
            
            for (int i = 1; i < ratings.length; ++i) {
                int new_slope = ratings[i] > ratings[i-1] ? 1 : ratings[i] < ratings[i-1] ? -1 : 0;
                if (old_slope > 0 && new_slope == 0 || old_slope < 0 && new_slope >= 0) {
                    candies += count(up) + count(down) + Math.max(up, down);
                    up = 0;
                    down = 0;
                }
                if (new_slope > 0) up++;
                if (new_slope < 0) down++;
                if (new_slope == 0) candies++;
                old_slope = new_slope;
            }
            
            candies += count(up) + count(down) + Math.max(up, down) + 1;
            
            return candies;
        }
    }
    

      

    reference:

    https://leetcode.com/problems/candy/solution/

    永远渴望,大智若愚(stay hungry, stay foolish)
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  • 原文地址:https://www.cnblogs.com/h-hkai/p/10202876.html
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